Estimate Instantaneous Velocity

[FritoTaco]

Homework Statement

If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.

(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.05 seconds

(iv) 0.01 seconds

(b) Estimate the instantaneous velocity when t = 2.

Homework Equations

Average Velocity =\dfrac{\triangle y}{\triangle t} (Change in position over change in time)

m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

The Attempt at a Solution

Set: t = 2 and t = 2 + h (let h be the time difference) \ne 0

t = 2 is y_{1},x_{1} and t = 2 + h is y_{2},x_{2}

Average Velocity = \dfrac{y(2+h)-y(2)}{(2+h)-2}

Now with y=48t-16t^2, we will replace t with 2 and 2 + h.

\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2)]}{h}

\dfrac{96+48h-16(4+4h+h^2)-96+64}{h}

\dfrac{48h-64-64h-16h^2+64}{h}

\dfrac{-16h^2-16h}{h}

\dfrac{h(-16h-16)}{h}

Average Velocity =-16h-16

Then I plug in (i) 0.5 into the equation like this: -16(0.5)-16=24
I do this for the rest of them and get correct answers according to the online "WebAssign" website. Now I'm stuck on (b) "Estimate the instantaneous velocity when t = 2." You can look at my picture, I tried plugging in values near 2 like 2.1 and 2.01, etc, into y=48t-16t^2 What am I doing wrong? So as I get closer to 2 I get closer to the slope of 31.999 which then you make an educated guess (by rounding) and get 32. But this is wrong. I don't know why? Also, I heard there is another method for finding average velocity, my book doesn't have it--anyone know it? Much appreciation.

 

[Mark44]

Homework Statement

If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.

(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.05 seconds

(iv) 0.01 seconds

(b) Estimate the instantaneous velocity when t = 2.

Homework Equations

Average Velocity =\dfrac{\triangle y}{\triangle t} (Change in position over change in time)

m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

The Attempt at a Solution

Set: t = 2 and t = 2 + h (let h be the time difference) \ne 0

t = 2 is y_{1},x_{1} and t = 2 + h is y_{2},x_{2}

Average Velocity = \dfrac{y(2+h)-y(2)}{(2+h)-2}

Now with y=48t-16t^2, we will replace t with 2 and 2 + h.

\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2)]}{h}

\dfrac{96+48h-16(4+4h+h^2)-96+64}{h}

\dfrac{48h-64-64h-16h^2+64}{h}

\dfrac{-16h^2-16h}{h}

\dfrac{h(-16h-16)}{h}

Average Velocity =-16h-16

Then I plug in (i) 0.5 into the equation like this: -16(0.5)-16=24
I do this for the rest of them and get correct answers according to the online "WebAssign" website. Now I'm stuck on (b) "Estimate the instantaneous velocity when t = 2." You can look at my picture, I tried plugging in values near 2 like 2.1 and 2.01, etc, into y=48t-16t^2 What am I doing wrong? So as I get closer to 2 I get closer to the slope of 31.999 which then you make an educated guess (by rounding) and get 32. But this is wrong. I don't know why? Also, I heard there is another method for finding average velocity, my book doesn't have it--anyone know it? Much appreciation.

For the instantaneous velocity near t = 2, look at the change in position divided by the change in time.

Look at $\frac{y(2 + h) - y(2)}{2 + h - 2}$, for some values of h that are close to 0, say h = .1, h = .01, h = .001.
You should get a value close to -16 ft/sec.

 

[FritoTaco]

My teacher also said use values like 0.1, 0.01, etc. Why? I don't get why we're using those values? Also, how do we make the equation where we can plug in those values? from using \dfrac{y(2+h)-y(2)}{(2+h)-(2)}

 

[Mark44]

My teacher also said use values like 0.1, 0.01, etc. Why? I don't get why we're using those values?

Because you want the time interval to be very short so as to get a closer estimate of the instantaneous velocity.

Also, how do we make the equation where we can plug in those values? from using \dfrac{y(2+h)-y(2)}{(2+h)-(2)}

You have the equation: $y(t) = 48t - 16t^2$

 

[FritoTaco]

Because you want the time interval to be very short so as to get a closer estimate of the instantaneous velocity.

This is where I don't understand. It says find instantaneous velocity when t = 2. Okay, so don't we want to use values close to 2? Like 2.1, 2.01. So if it was like, "find instantaneous velocity closer to when t = 5", would I still use 0.1, 0.01, etc? Because that means we are 0.1 from 5 or 0.01 from 5?

You have the equation: y(t)=48t−16t2

When I plug 48(t)-16(t)^2 into \dfrac{y(2+h)-y(2)}{(2+h)-2}, I don't understand how to do it. Am I actually plugging in 48(t)-16(t)^2 into the y values of the other equation? Something like this:

\dfrac{48t-16t^2(2+h)-48t-16t^2(2)}{2+h-2}? Then distribute and simplify it down more?

 

[Mark44]

This is where I don't understand. It says find instantaneous velocity when t = 2. Okay, so don't we want to use values close to 2? Like 2.1, 2.01. So if it was like, "find instantaneous velocity closer to when t = 5", would I still use 0.1, 0.01, etc? Because that means we are 0.1 from 5 or 0.01 from 5?

Yes. If h = .01, then 2 + h = 2 + .01 = 2.01

When I plug 48(t)-16(t)^2 into \dfrac{y(2+h)-y(2)}{(2+h)-2}, I don't understand how to do it. Am I actually plugging in 48(t)-16(t)^2 into the y values of the other equation? Something like this:
\dfrac{48t-16t^2(2+h)-48t-16t^2(2)}{2+h-2}? Then distribute and simplify it down more?

No, you are "plugging in" 2 + h and 2, respectively, into your height function y(t) = 48t - 16t²
For y(2 + h), replace t in your formula with 2 + h.
For y(2), replace t with 2.
Then simplify the result.

 

[FritoTaco]

No, you are "plugging in" 2 + h and 2, respectively, into your height function y(t) = 48t - 16t2
For y(2 + h), replace t in your formula with 2 + h.
For y(2), replace t with 2.
Then simplify the result.

48(2+h)-16(2)^2

96+48h-64

48h-32

Am I doing this right?

 

[Mark44]

48(2+h)-16(2)^2

I don't know what you're doing above.

96+48h-64

48h-32

Am I doing this right?

No. This is what you should be evaluating: $\frac{y(2+h)-y(2)}{(2+h)-2}$, using $y(t) = 48t - 16t^2$. The quotient here gives you the average speed between time t = 2 + h and time t = 2.
Edit: Fixed a typo I had in the y(t) formula.

 

[FritoTaco]

No. This is what you should be evaluating: y(2+h)−y(2)(2+h)−2y(2+h)−y(2)(2+h)−2\frac{y(2+h)-y(2)}{(2+h)-2}, using y(t)=48t−32t2y(t)=48t−32t2y(t) = 48t - 32t^2. The quotient here gives you the average speed between time t = 2 + h and time t = 2.

How did you get y(t)=48t-32t^2?

Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the 48h - 32?

 

[Mark44]

How did you get y(t)=48t-32t^2?

I miswrote 32t² instead of 16t². I have fixed my typo.

Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the 48h - 32?

Whatever you did isn't right. y(2 + h) gives you two terms, and y(2) also gives you two terms. Also, you have 2 + h - 2 in the denominator.

 

[Mark44]

Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the 48h - 32?

Let's back up a few steps. You need to evaluate $\frac{y(2 + h) - y(2)}{2 + h - 2}$, using $y(t) = 48t - 16t^2$.

What do you get for y(2 + h)?
What do you get for y(2)?

 

[FritoTaco]

Am I placing 48t-16t^2 into the y part of both y(2+h) and y(2)?

Like this maybe:

\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2]}{(2+h)-2}

\dfrac{96+48h-16(h^2+4h+4)-96+64}{(2+h)-2}

\dfrac{96+48h-16h^2-64h-64-96+64}{(2+h)-2}

\dfrac{-16h^2-16h}{(2+h)-2}

I'm pretty sure something isn't right but I gave it a go here.

 

[Mark44]

Am I placing 48t-16t^2 into the y part of both y(2+h) and y(2)?

Like this maybe:

\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2]}{(2+h)-2}

\dfrac{96+48h-16(h^2+4h+4)-96+64}{(2+h)-2}

\dfrac{96+48h-16h^2-64h-64-96+64}{(2+h)-2}

\dfrac{-16h^2-16h}{(2+h)-2}

I'm pretty sure something isn't right but I gave it a go here.

This is correct, but you should simplify the denominator.

When you've done that, estimate the velocity at a few times close to 2 sec., such as 2.1 (h = .1), 2.01 (h = .01), and however more you need to do.

 

[FritoTaco]

This is correct, but you should simplify the denominator.

When you've done that, estimate the velocity at a few times close to 2 sec., such as 2.1 (h = .1), 2.01 (h = .01), and however more you need to do.

Will it just be \dfrac{-16h^2-16h}{h}?

 

[Mark44]

Will it just be \dfrac{-16h^2-16h}{h}?

And simplfy that...

 

[FritoTaco]

Like this:

\dfrac{-16h(h+1)}{h} Factor

16(h+1) Cancel out the h

 

[Mark44]

Like this:

\dfrac{-16h(h+1)}{h} Factor

16(h+1) Cancel out the h

Where did the minus sign go?

When you get that straigtened out, pick a few (small) values of h, as described in parts i) through iv) of the first post, and then estimate the velocity at t = 2 seconds.

 

[FritoTaco]

Oh, I forgot about that. It should've been there as -16(h+1)

i.)0.5

-16(0.5+1) = -24 ft/s​

ii.)0.1

-16(0.1+1) = -17.6 ft/s​

iii.)0.05

-16(0.05+1) = -16.8 ft/s​

iv.)0.01

-16(0.01+1) = 1-6.16 ft/s​

b.) Estimate the instantaneous velocity when t = 2.

-16(0.1+1) = -17.6 ft/s

-16(0.01+1) = -16.16 ft/s

-16(0.001+1) = -16.08 ft/s

Estimate: -16 ft/s.

I entered that in and got it right! Thank you very much for your help!

 

[Mark44]

iv.)0.01

-16(0.01+1) = 1-6.16 ft/s

Surely, you mean -16.16 ft/sec

b.) Estimate the instantaneous velocity when t = 2.

-16(0.1+1) = -17.6 ft/s

-16(0.01+1) = -16.16 ft/s

-16(0.001+1) = -16.08 ft/s

Estimate: -16 ft/s.

I entered that in and got it right! Thank you very much for your help!

 

[FritoTaco]

Oh, oops. Yeah, -16.16ft/s.